and that TSFC increases linearly with specific thrust.MAE 113 HW1 solution-2. d) explain the trends You're on your own here. Say something about how TSFC is always higher when the inlet velocity is higher.nb
3
The blue line is the V0 = 0 line and the green line is the V0 = 500 ft ê s line.
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.

It has radius r0 . We can combine these two equations to get ⁄Fs = ° ° IM out .M inM = .MAE 113 HW1 solution-2.M in M
dMs dt
by assuming conservation of mass (since there is no sink or source).M in M
Where F is the force required to keep the pipe in place and A is the area of the pipe.P2 L A
1 gc
c
These equations are wrong. ° ° ° ° because that would imply that if M in > M out. This means that we have defined our sum of forces to be in a different direction in each of the equations. velocity V1 . M in > M out implies that air is moving from back to front. Instead. ⁄Fs = The sum of the forces is ⁄Fs = -F + HP1 . the force would be postive. the first part of this equation does not make sense.I r M F
r 2
0
with uniform pressure P2 .P2 L A
c
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. Use the conservation of mass and momentum equations to show that the force necessary to hold the pipe in place can be written as F = p r2 JP1 . and pressure P 1 .
° ° IM out .P2 + 0 Start with equation 2.M out. Force should not be proportional to M in . which should decrease the force required to keep the pipe steady.3 Consider the flow shown in Figure P2.M outM + HP1 . the equation should read: ° ° 1 F = g IM out .P2 L A ° ° 1 F = g IM in .F + HP1 . The fluid leaves with a velocity V2 = Vmax B1 . ° ° However. However.2.P2 L A
1 gc
= 0.M inM + HP1 .20 ⁄Fs =
1 gc
2 r V1 N 3 gc
I
dMs dt
° ° + M out .nb
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2.